We know that because the integer $n$ is an integer, both $n-1$ and $n+1$ have to both be integers. Let us note that a square root of a positive integer that is not a square added to another square root of a positive integer that is not a square would be an irrational number because it cannot be written as $\frac{p}{q}$ such that $p$ and $q$ are integers. In order for $\sqrt{n-1}$ to not be complex, $n\geq 1.$ In order for $\sqrt{n+1}$ to not be complex, $n\geq -1.$ In order for $\sqrt{n-1}$ and $\sqrt{n+1}$ to both be rational numbers, we need $n-1$ and $n+1$ to both be squares. We let $a^2=n+1$ and $b^2=n-1$. Because $n-1$ and $n+1$ both have to be perfect squares, $a$ and $b$ are both integers We know that the $a^2-b^2=2 \Longrightarrow (a+b)(a-b)=2.$ Let us remeber that $a>b,$ thus both $(a+b)$ and $(a-b)$ both have to be positive integers. The only possible way that could every occur would be if $$a-b=1$$and $$a+b=2.$$Solving this system of equations we get that $a=1.5$ and $b=.5$ We had already established that in order for this inequlity to be true is to have $a$ and $b$ both be integers. Thus, this inequality is false.

Suppose that there is an integer $n$ such that this is rational. Note that $\sqrt{n-1}, \sqrt{n+1}$ are both algebraic integers, so their sum is also an algebraic integer, but by the rational root theorem, the intersection of the set of algebraic integers and the set of rational numbers is an integer, so $\sqrt{n-1} + \sqrt{n+1}$ is an integer. We have $2\sqrt{n-1}\leq \sqrt{n-1} + \sqrt{n+1}\leq 2\sqrt{n+1}$, and since $2(\sqrt{n+1} - \sqrt{n-1}) = \frac{4}{\sqrt{n+1} + \sqrt{n-1}}$, the function $f(n) = 2(\sqrt{n+1} + \sqrt{n-1})$ is monotonically decreasing on $(1, \infty)$ (clearly $n > 1$ because then the number would be complex) so the interval shrinks, and letting $g(n) = \sqrt{n-1} + \sqrt{n+1}$, since $g(1) = \sqrt{2}$, not an integer, $g(2) = 1 + \sqrt{3}$, not an integer, $g(3) = 2 + \sqrt{2}$, not an integer, $g(4) = \sqrt{3} + \sqrt{5}$, not an integer, and $f(5) = 2(\sqrt{6} - 2) < 1$ as $2.4 < \sqrt{6} < 2.5$. Thus there is no such integer $n$.